This perhaps is not a math problem. I donot know if it fits in here. But it confuses me a lot these two days.
I have wiki it. But it does not work for me. I will give some easy "proofs" to explain my confusion.

Prove that every set has the same cardinality with itself.

I prove it by contradiction. Suppose this statement is not true, then we pick one from what?... The elements fail in this statement cannot form a set?(How do I test whether it is a set? Is "pick" not a map which defined by set-theory ?). Even this is not a set, we still can pick one, right?

Confusion comes from induction.

General speaking, I donot encounter troubles use induction. But many authors like the following manner which I think is equivalent to induction: Suppose statement $S$ is not true for all natural numbers. By well ordered property of natural numbers, we can find a smallest number $n$ such that $S(n)$ is not true. Then....
Until here no problems. But what if the elements in $S(n)$ is not a set. For example, we want to show statement about all finitely generated $R$-module is true. When $R$ runs, module structure runs, can I say that "since $S(n)$ is not true, say $(M,T)$ is false in $S$,(where $M$ is a $T$ module generated by $n$ elements)" ? But here we can fix a ring $R$ first, then continue.

I think there are several inconsistencies in my argument.

Next is a wrong proof but not wrong too much.

Show every field has an algebraic closure.

proof. Say $k$ a field. Consider $\Sigma=\{L|L \text{ is an algebraic extension of }k\}$. Apply zorn's lemma, easily find a maximal element in $\Sigma$, then this element is a algebraic closure of $k$. But be carefully, we do not test whether $\Sigma$ is a set!

All right. I am sorry, I cannot figure out what my confusions.

I am not familiar with set-theory.

What exactly is a statement?
How am I convinced myself that I have gave a right proof?

What is a fast way to convince a thing is a set for people having only naive set-theory foundation? Or how do we avoid this(not need to judge whether a class is a set) ?(This will be a primary question in this post).

1 Answer
1

The following is very far from being a complete answer. My first approximation to a response is to advise you not to worry about it. A second approximation is to claim that the standard set theory ZFC (Zermelo-Fraenkel with Choice) provides all the set construction tools that one ever needs.

That claim is in fact false. There are some difficulties in setting up Category Theory in a ZFC framework. And if you are interested in "large cardinals," (strongly inaccessible, measurable, and so on), again there are some problems. But ZFC is certainly strong enough for all of the ordinary mathematical constructions.

This includes everything that you mentioned, and far far more. However, on occasion one has to make some modification of a "naive" construction in order to avoid bumping into "the set of all sets," which, as you know, produces contradictions if it is assumed to behave like "real" sets.

Now to (some) details. Your first question is about proving that every set has the same cardinality as itself. The way Cantor, and other early set theorists, defined the cardinality of a set was to define two sets to be equipollent if there was a bijection between them. Then they defined the cardinality of a set $X$ to be the "set" of all sets equipollent with $X$, and called this "set" a cardinal number.

Of course that leads to immediate problems. So the notion of cardinal number had to be redefined. There are various redefinitions, but in most presentations one first of all defines ordinal, and then defines a cardinal as an initial ordinal. I really cannot give the details, cannot write a chapter on set theory as a response! But the details are fully worked out in many books.

It is then trivial to show that a set has the same cardinality as itself. The cardinality of set $X$ is by definition the initial ordinal $\kappa$ such that there is a bijection between $X$ and $\kappa$. This initial ordinal is the same for $X$ as it is for $\dots$ $X$!

Now let's look at the question you asked about algebraic closure. As you are obviously aware, the definition of $\Sigma$ is "wrong," or at least problematic, since $\Sigma$, as defined, is too large to be a set. So we must change the definition, in order to avoid dealing with what, in old-fashioned language, used to be called inconsistent magnitudes.

The fix is easy. Informally, we know that the algebraic closure of an infinite field $k$ will have cardinality the same as the cardinality of $k$. So we can pick a fixed set $V$, that contains $k$, and that has cardinality the next cardinal after the cardinality of the field $k$, with the obvious modification for finite fields.

We can then use almost the same definition as the one that you gave, except that we consider only fields $L$ whose underlying set is a subset of $V$. The remaining details of the argument remain essentially unchanged. (I chose "the next cardinal" to allow plenty of room for the standard construction. One can do it cheaper, but then imitating the naive proof takes a bit more care.)

Often, if we know that an informal idea based on naive set theory has a standard obvious fix of the type described above for algebraic closure, we can afford to be sloppy. That is, we can give a naive set theory definition, knowing that we could fix it if challenged.

Summary: Every issue that you raised was dealt with, I think satisfactorily, quite a few years ago. So not worrying about the matter is a sensible attitude, though perhaps not an intellectually defensible one.

If you continue to be concerned about possible set-theoretic inconsistencies in the mathematics that you do, you will need to work through at least the first few chapters of a "modern" set theory book, say from the $1970$s or later. After that, for most fields of mathematics, you need not worry about set theory again. There are some exceptions. Occasionally, particularly in set-theoretic topology, or related subjects, questions come up that really do involve deep set-theoretic issues.